7c+6=(c-4)2c

Solution for 7c+6=(c-4)2c equation:

7c+6=(c-4)2c

We move all terms to the left:

7c+6-((c-4)2c)=0


We calculate terms in parentheses: -((c-4)2c), so:

(c-4)2c

We multiply parentheses

2c^2-8c

Back to the equation:

-(2c^2-8c)


We get rid of parentheses

-2c^2+7c+8c+6=0

We add all the numbers together, and all the variables

-2c^2+15c+6=0

a = -2; b = 15; c = +6;
Δ = b2-4ac
Δ = 152-4·(-2)·6
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{273}}{2*-2}=\frac{-15-\sqrt{273}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{273}}{2*-2}=\frac{-15+\sqrt{273}}{-4}
$